Question

if 16x4-8x3+x2-8x+4 is divided by 2x-1 then the remainder is​

Answer 1

Answer:

g(x)=0=2x-1

     =1=2x

     =1/2=x

by using remainder theorem as we know that when p(x) divided by 2x-1 then remainder =p(1/2)

$p(\frac{1}{2} ) = 16*\frac{1}{2}^{4} -8*\frac{1}{2} ^{3} +\frac{1}{2} ^{2} -8*\frac{1}{2} +4\\=1- 1 +\frac{1}{4} -4+4\\=\frac{1}{4}$

remainder=1/4

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Step-by-step explanation:

Answer 2

Answer:

If $16 x^4-8 x^3+x^2-8 x+4$ is divided by $2x-1$ the remainder is 1/8 .

Step-by-step explanation:

Given :

The given polynomial is $16 x^4-8 x^3+x^2-8 x+4$ and the divisor is $2x-1$ .

To solve :

The remainder theorem states that the remainder will be P(a) when a polynomial P(x) is divided by $x-a$,   .

Remainder Theorem is a method which divides polynomials according to Euclidean geometry. This theorem describes that when a polynomial P(x) is divided by a factor (x - a), which isn't really an element of the polynomial, a smaller polynomial is produced along with a remainder.

In the given question, we have $2x-1$ .

Put $2x-1=0$

So the value of $x=\frac{1}{2}$

Now $P(X)=16 x^4-8 x^3+x^2-8 x+4$

Put the values in the equation.

$P(\frac{1}{2})=16 (\frac{1}{2})^{4} -8(\frac{1}{2})^{3} +(\frac{1}{2})^{3} -8(\frac{1}{2} )+4\\ P(\frac{1}{2})=\frac{1}{8}$

Therefore, the remainer is $\frac{1}{8}$ .

To know more about "remainder theorem"

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